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5.92 g of sodium oxalate is reacted with 5.92 of calcium chloride follow the format used in the image provided.

5.92 g of sodium oxalate is reacted with 5.92 of calcium chloride follow the format used in the image provided.

Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter3: Chemical Reactions
Section: Chapter Questions
Problem 82IL
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Directions: Write the balanced equation for each of the following situations. SHOW ALL OF YOUR 
WORK ON ATTACHED PAGES, OR IT WILL NOT BE ACCEPTED. In addition, list the reaction type.
YOU MUST TELL THE AMOUNTS OF EVERY SUBSTANCE THAT REMAINS IN THE 
CONTAINER AT THE END OF THE REACTION. ASSUME THAT ALL REACTIONS 
GO TO COMPLETION.
If only STOICHIOMETRY, tell how much of the excess reactant is used!!!!
 Reaction Type
a. Combination Reaction
b. Decomposition Reaction
c. Single Displacement / THIS IS ONE TYPE OF Oxidation Reduction Reaction 
d. Precipitation Reaction
e. Gaseous Reaction
f. Neutralization Reaction
g. Combustion Reaction

5.92 g of sodium oxalate is reacted with 5.92 of calcium chloride

follow the format used in the image provided.

2. N₂ (g) + 3 H₂ (g) 2 NH3 (g) (This is LIMITING REACTANT: N₂ is the Limiting Reactant) Grams of Product (List the Product and the Amount): 0.75168 g NH3 Reaction Type: Combination or Single Displacement (which is also called Oxidation-Reduction) 1g N₂ ? g NH3 = 61.802 cg N₂ x LR 1 mol N₂ x 2 mol NH3 x 28.02 g N₂ 1 mol N₂ 17.04 g NH3 = 1 mol NH3 1 x 10²2 cg N₂ ? g NH3 = 61.802 cg H₂ x How much N₂ remains in the vessel? 1g H₂ 1 x 10² g H₂ X X X 1 mol H₂ 2.02 g H₂ X 2 mol NH3 x 17.04 g NH3 = 3 mol H₂ 1 mol NH3 0.75168 g NH3 ******* THEORETICAL YIELD You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. ? g H₂ USED = 61.802 cg N₂ X 1g N₂ 1 mol N₂ x 28.02 g N₂ 1 x 10² cg N₂ Amount of H₂ Remaining in the Container = H₂ amount given - H₂ amount USED= 0.61802 g H₂ GIVEN - 0.13366 g H₂ USED = 0.48436 g of H₂--LEFT OVER = EXCESS 3 mol H₂ x 2.02 g H₂ = 1 mol N₂ 1 mol H₂ 3.3756 g NH3 0.13366 g H₂
Transcribed Image Text:2. N₂ (g) + 3 H₂ (g) 2 NH3 (g) (This is LIMITING REACTANT: N₂ is the Limiting Reactant) Grams of Product (List the Product and the Amount): 0.75168 g NH3 Reaction Type: Combination or Single Displacement (which is also called Oxidation-Reduction) 1g N₂ ? g NH3 = 61.802 cg N₂ x LR 1 mol N₂ x 2 mol NH3 x 28.02 g N₂ 1 mol N₂ 17.04 g NH3 = 1 mol NH3 1 x 10²2 cg N₂ ? g NH3 = 61.802 cg H₂ x How much N₂ remains in the vessel? 1g H₂ 1 x 10² g H₂ X X X 1 mol H₂ 2.02 g H₂ X 2 mol NH3 x 17.04 g NH3 = 3 mol H₂ 1 mol NH3 0.75168 g NH3 ******* THEORETICAL YIELD You will use the LIMITING REACTANT and determine how much H₂ was USED in the RXN. ? g H₂ USED = 61.802 cg N₂ X 1g N₂ 1 mol N₂ x 28.02 g N₂ 1 x 10² cg N₂ Amount of H₂ Remaining in the Container = H₂ amount given - H₂ amount USED= 0.61802 g H₂ GIVEN - 0.13366 g H₂ USED = 0.48436 g of H₂--LEFT OVER = EXCESS 3 mol H₂ x 2.02 g H₂ = 1 mol N₂ 1 mol H₂ 3.3756 g NH3 0.13366 g H₂
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